u1 = (m1-m2) •v/(m1+m2) – (m-16m) •250/(m+16m) = - 220.6 m/s,
Minus indicates that the proton is returned back due to the collision.
m1•v = m1•u1+m2•u2
u2 = (v+u1)m1/ m2 = 250 –(-220.6) •m/16•m = 29.4 m/s
in the direction of the initial direction of the proton.
A proton that has a mass m and is moving at 250 m/s in the + direction undergoes a head-on elastic collision with a stationary oxygen nucleus of mass 16m. Find the velocities of the proton and the oxygen nucleus after the collision
i know i have to consider that both momentum and KE are conserved but i don't know what to do (how to do it)
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